The horizontal component of the earth’s magnetic field at a certain place is 2.6×10^-5T and the angle of dip is 60°. What is the magnetic field of the earth at this location?

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Detailed Solution

$\begin{array}{l} \cos θ = \frac{H}{B} \Rightarrow B = \frac{H}{\cos 60} = \frac{2 .6 \times 10^{- 5}}{\frac{1}{2}} \\ = 5 .2 \times 10^{- 5} T \end{array}$

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